Question

The least positive integral value of $\lambda $ for which $f(x) = \dfrac{{3{x^3}}}{2} + \dfrac{{\lambda {x^2}}}{3} + x + 7$ has a point of maxima is ?
A) $6$
B) $7$
C) $8$
D) $10$

Answer 1

Hint: We can find the first derivative of the given function. Then find the discriminant. The maxima is given by the solution of discriminant greater than zero. Thus we get the minimum value for $\lambda $. So we can find the nearest integer to it.

Formula used:
Discriminant of a quadratic equation of the form $a{x^2} + bx + c$ is given by $D = {b^2} - 4ac$.

Complete step-by-step answer:
Given the function, $f(x) = \dfrac{{3{x^3}}}{2} + \dfrac{{\lambda {x^2}}}{3} + x + 7$
We have to find the least positive integral value of $\lambda $ for which this function attains maximum.
Differentiating both sides with respect to $x$ we get,
$f'(x) = \dfrac{{9{x^2}}}{2} + \dfrac{{2\lambda x}}{3} + 1$
Making denominators common we get,
$\Rightarrow$ $f'(x) = \dfrac{{27{x^2}}}{6} + \dfrac{{4\lambda x}}{6} + \dfrac{6}{6}$
Taking $\dfrac{1}{6}$ common outside we get,
$\Rightarrow$ $f'(x) = \dfrac{1}{6}(27{x^2} + 4\lambda x + 6)$
For the function to have maximum, the discriminant of its derivative must be greater than zero.
For a quadratic equation
$a{x^2} + bx + c = 0$
Discriminant is given by,
D=$b^2$-$4ac$
Here,
$\Rightarrow$ $D = {(4\lambda )^2} - 4 \times 27 \times 6 = 16{\lambda ^2} - 24 \times 27$
$\Rightarrow$ $D > 0 \Rightarrow 16{\lambda ^2} - 24 \times 27 > 0$
Dividing both sides we have,
$\Rightarrow$ $2{\lambda ^2} - 3 \times 27 > 0$
$ \Rightarrow 2{\lambda ^2} - 81 > 0$
This gives,
$2{\lambda ^2} > 81 \Rightarrow {\lambda ^2} > \dfrac{{81}}{2} \Rightarrow \left| \lambda \right| > \dfrac{9}{{\sqrt 2 }}$
We know that
$\Rightarrow$ $\dfrac{9}{{\sqrt 2 }} \sim 6.36$
So the least integer value is given by the next integer which is equal to seven.
Therefore the answer is option B.

Note: Solving the discriminant equation, we got $\left| \lambda \right| = \dfrac{9}{{\sqrt 2 }}$. It has both positive and negative values. We neglect the negative value since the question is asked for the least positive value.
The question is to find the maxima. But solving the discriminant we got the minimum value for $\lambda $. So we chose the integer greater than the actual value. Instead if we got the maximum value of $\lambda $, we would have chosen the integer less than it which is equal to six.